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If AD is 6 and ADC is a right angle, what is the area of

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Re: Area of triangular region [#permalink]

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15 Dec 2013, 23:21
AccipiterQ wrote:

I thought if you dropped a line down from a triangle vertex and it formed a right angle on the opposite side then that line bisected the side? So in this case if you know what BD is then you know what DC is?

To figure out whether it holds, why don't you try drawing some extreme figures, say, something like this:
Attachment:

Ques3.jpg [ 3.67 KiB | Viewed 1345 times ]

Will this be true in this case?
When will it be true? When the triangle is equilateral, sure. Also when the triangle is isosceles if the equal sides form the angle from which the altitude is dropped.

Don't put your faith in the figure given. It may be just one of the many possibilities or may be somewhat misleading.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 16974 [0], given: 230 Director Affiliations: CrackVerbal Joined: 03 Oct 2013 Posts: 512 Kudos [?]: 553 [0], given: 6 Location: India GMAT 1: 780 Q51 V46 Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink] Show Tags 15 Dec 2013, 23:50 amitjash wrote: Attachment: Triangle.jpg If AD is 6 and ADC is a right angle, what is the area of triangular region ABC? (1) Angle ABD = 60° (2) AC = 12 Statement I is insufficient The left triangle becomes a 90 30 60 triangle however we don't know anything about the right triangle Statement II is insufficient The right triangle becomes a 90 30 60 triangle however we don't know anything about the left triangle Combining is sufficient We know both the triangles are 90 30 60 with same sides. Hence Answer is C Pushpinder Gill _________________ Enroll for our GMAT Trial Course here - http://gmatonline.crackverbal.com/ Learn all PS and DS strategies here- http://gmatonline.crackverbal.com/p/mastering-quant-on-gmat For more info on GMAT and MBA, follow us on @AskCrackVerbal Kudos [?]: 553 [0], given: 6 Intern Joined: 17 Jun 2013 Posts: 5 Kudos [?]: 3 [0], given: 29 Schools: ISB '14 Re: Area of triangular region [#permalink] Show Tags 24 Jan 2014, 22:10 Hi Bunuel, I have a question. Can I not imagine that, in a right angled triangle if one of the side of a triangle bears x square root 3 as its length can I not imagine that it's a 30-60-90? If not why. Please help. Thanks Suneel Kudos [?]: 3 [0], given: 29 Math Expert Joined: 02 Sep 2009 Posts: 41717 Kudos [?]: 124948 [0], given: 12079 Re: Area of triangular region [#permalink] Show Tags 25 Jan 2014, 03:39 dasikasuneel wrote: Hi Bunuel, I have a question. Can I not imagine that, in a right angled triangle if one of the side of a triangle bears x square root 3 as its length can I not imagine that it's a 30-60-90? If not why. Please help. Thanks Suneel What do you mean by "imagine"? How does the length of one side define the angles? _________________ Kudos [?]: 124948 [0], given: 12079 Intern Joined: 30 Jan 2014 Posts: 1 Kudos [?]: [0], given: 5 Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink] Show Tags 06 Feb 2014, 11:28 Bunuel wrote: manimgoindowndown wrote: Hey I had the same question as the last poster. How do we know that BD and DC are of the same length? or that angle BAC has been bisected? They are not equal: $$BD=\frac{6}{\sqrt{3}}$$ (from the first statement) and $$AD=6*\sqrt{3}$$ (from the second statement). I think answer is E, since it is not given that BDC points are collinear. Kudos [?]: [0], given: 5 Math Expert Joined: 02 Sep 2009 Posts: 41717 Kudos [?]: 124948 [1], given: 12079 Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink] Show Tags 07 Feb 2014, 05:27 1 This post received KUDOS Expert's post virendrasd wrote: Bunuel wrote: manimgoindowndown wrote: Hey I had the same question as the last poster. How do we know that BD and DC are of the same length? or that angle BAC has been bisected? They are not equal: $$BD=\frac{6}{\sqrt{3}}$$ (from the first statement) and $$AD=6*\sqrt{3}$$ (from the second statement). I think answer is E, since it is not given that BDC points are collinear. That's not correct. OG13, page 272: A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2). Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight. You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees. All figures lie in a plane unless otherwise indicated. OG13, page 150: Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated. Hope it helps. _________________ Kudos [?]: 124948 [1], given: 12079 Senior Manager Joined: 06 Aug 2011 Posts: 393 Kudos [?]: 229 [0], given: 82 Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink] Show Tags 18 Mar 2014, 01:11 what if question wud have been like this.. If AD is 6, and ADC is a right angle, what is the area of equilateral triangle ABC? _________________ Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score ! Kudos [?]: 229 [0], given: 82 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7617 Kudos [?]: 16974 [0], given: 230 Location: Pune, India Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink] Show Tags 18 Mar 2014, 08:51 sanjoo wrote: what if question wud have been like this.. If AD is 6, and ADC is a right angle, what is the area of equilateral triangle ABC? Then statement 1 would have been irrelevant and statement 2 would have been incorrect. If side of an equilateral triangle is 12, the altitude would be $$(\sqrt{3}/2)*12 = 6*\sqrt{3}$$ But AD is given to be 6. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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12 Apr 2015, 15:31
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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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12 Apr 2015, 17:23
Main take away from this question is:
Figures are not drawn to scale.

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what is the area of triangular region [#permalink]

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08 Jun 2015, 10:29
If AD is 6 , and ADC is a right angle, what is the area of triangular region ABC?
(1) Angle ABD = 60°
(2) AC = 12
Attachments

triangleabc.jpg [ 7.64 KiB | Viewed 766 times ]

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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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08 Jun 2015, 10:38
Mechmeera wrote:
If AD is 6 , and ADC is a right angle, what is the area of triangular region ABC?
(1) Angle ABD = 60°
(2) AC = 12

Merging topics.

Please refer to the discussion on pages 1 and 2.
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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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25 Oct 2016, 06:46
amitjash wrote:
Attachment:
Triangle.jpg
If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°
(2) AC = 12

I tried not to solve everything as it is a DS question...
Area is equal to base x height / 2
AD is height, BC is the base. We need to find the value of BC.
1. since B is 60 degree, A in ABD is 30.
we have 1 leg, and we know the property of 30-60-90 right triangle
we can find BD.
but since we don't know whether AD divides BC in two equal parts, then we can't tell exactly whether we can find BC.

2. AC = 12. We can find DC. we are faced with the same problem - can't find BD.

1+2
we have BD and we have DC. we can find the answer.

C

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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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14 Aug 2017, 00:28
AccipiterQ wrote:
Bunuel wrote:
amitjash wrote:
If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?
1. Angle ABD=60
2. AC=12

Can someone explain???

There should be a diagram attached:
Attachment:
2cr0lsz.jpg

Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio $$1:\sqrt{3}:2$$ --> as AD=6 (larger leg opposite 60 degrees angle) then $$BD=\frac{6}{\sqrt{3}}$$ (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

I thought if you dropped a line down from a triangle vertex and it formed a right angle on the opposite side then that line bisected the side? So in this case if you know what BD is then you know what DC is?

I also assumed a similar thing . But I believe a perpendicular only bisects the angle at the vertex IF the perpendicular bisects the opposite side .. and vice versa . So we cannot assume the perpendicular is an angle / side bisector unless it was explicitly mentioned in the stem

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Re: If AD is 6 and ADC is a right angle, what is the area of   [#permalink] 14 Aug 2017, 00:28

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